C: printf() and putchar() questions

Question

I was reading K&R book and wanted to test out printf() and putchar() functions in ways that I never tried. I encountered several unexpected events and would like to hear from more experienced programmers why that happens.

char c;
while((c = getchar()) != EOF) {
//put char(c);
    printf("%d your character was.\n", c);
}

How would one get EOF (end of file) in the input stream (getchar() or scanf() functions)? Unexpected key that is not recognized by getchar()/scanf() function could produce it?
In my book, it says that c has to be an integer, because it needs to store EOF and the variable has to be big enough to hold any possible char that EOF can hold. This doesn't make sense for me, because EOF is a constant integer with a value of -1, which even char can store. Can anyone clarify what was meant by this?
What happens when I send "hello" or 'hello' to putchar() function? It expects to get an integer, but returns weird output, such as EE or oo, if I send the latter string or char sequence.
Why when I use printf() function that is written above I get two outputs? One is the one I entered and the other one is integer, which in ASCII is end of line. Does it produce the second output, because I press enter, which it assumes to be the second character?

Thanks.

EOF usually is "-1", getchar() returns it when read() from terminal returns 0. You can't use char for getchar because then in case of printing character with code 255 (for example, 'Ъ' in KOI8-R) you'll get a false EOF. To hide input characters you should properly setup your terminal (turn off echo). — Eddy_Em
by saying false EOF, you mean that the character is an integer larger than 1 byte and can be recognized by getchar(), if the variable's range is set to be more than 1 byte? — Jonas Hoffmann
'hello' is not a valid string, it possibly results in undefined behaviour. Something between '' must only represent a single character. — Bernhard Barker
By saying "false EOF" I mean that you enter a character but your application will think that that is EOF. Any letter == unsigned char, so, you need at least 2 bytes to send EOF! — Eddy_Em
I agree that any letter is an unsigned char, but char's range is also negative numbers. How is it that EOF need 2 bytes if it's -1? What characters can a person input to make the application think that it's EOF? — Jonas Hoffmann

Guillaume Guillaume · Accepted Answer · 2013-06-24T10:20:15

on Linux, you can send it with Ctrl+d
you need an int, otherwise you can't make the difference between EOF and the last possible character (0xFFFF is not the same than 0x00FF)
putchar wants a character, not a string, if you're trying to give it a string, it'll print a part of the string address
you only get one output: the ascii value of the character you entered, the other "input" is what you typed in the terminal

edit - more details about 2

You need an int, because getchar can returns both a character of value -1 (0x00FF) and an integer of value -1 (0xFFFF), they don't have the same meaning: the character of value -1 is a valid character (for instance, it's ÿ in latin-1) while the integer of value -1 is EOF in this context.

Here's a simple program that shows the difference:

#include <stdio.h>

int main(int argc, char ** argv) {
    {
        char c = 0xFF; /* this is a valid char */
        if (c == EOF) printf("wrong end of file detection\n");
    }
    {
        int c = 0xFF; /* this is a valid char */
        if (c == EOF) printf("wrong end of file detection\n");
    }
}

The first test succeeds because 0xFF == -1 for char, while the second tests fails because 0x00FF != -1 for int.

I hope that makes it a bit clearer.

C: printf() and putchar() questions

2 Answers